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shusha [124]
3 years ago
5

The options are:

Mathematics
1 answer:
kati45 [8]3 years ago
8 0

Answer:

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...................

..............

0

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Given that PQRS is a square, find x and y.
Dovator [93]
Every angle of a square equals 90 degrees.

2x - 4 = 90              3y + 6 = 90
2x = 90 + 4              3y = 90 - 6
2x = 94                    3y = 84
x = 94/2                     y = 84/3
x = 47                        y = 28

so x = 47 and y = 28
4 0
4 years ago
Read 2 more answers
Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$
Anna [14]

Answer:

\large\boxed{1\dfrac{1}{3}\ u^2}

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}

6 0
3 years ago
To find the sum of 4 and -5.
Katen [24]

Answer:

-1

Step-by-step explanation:

-5 + 4 = 1

5 0
4 years ago
A container is the shape of an inverted right circular cone has a radius of 4.00 inches at the top and a height of 5.00 inches.
Len [333]

The rate at which the water from the container is being drained is 24 inches per second.

Given radius of right circular cone 4 inches .height being 5 inches, height of water is 2 inches and rate at which surface area is falling is 2 inches per second.

Looking at the image we can use similar triangle propert to derive the relationship:

r/R=h/H

where dh/dt=2.

Thus r/5=2/5

r=2  inches

Now from r/R=h/H

we have to  write with initial values of cone and differentiate:

r/5=h/5

5r=5h

differentiating with respect to t

5 dr/dt=5 dh/dt

dh/dt is given as 2

5 dr/dt=5*-2

dr/dt=-2

Volume of cone is 1/3 πr^{2} h

We can find the rate at which the water is to be drained by using partial differentiation on the volume equation.

Thus

dv/dt=1/3 π(2rh*dr/dt)+(r^{2}*dh/dt)

Putting the values which are given and calculated we get

dv/dt=1/3π(2*2*2*2)+(4*2)

=1/3*3.14*(16+8)

=3.14*24/3.14

=24 inches per second

Hence the rate at which the water is drained from the container is 24 inches per second.

Learn more about differentaiation at brainly.com/question/954654

#SPJ4

5 0
2 years ago
If we were in the core of the galaxy...
Svetllana [295]

Answer:

B

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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